Trigonometry implicit differentiation

Question

If ${y}+\sin(y) = \cos(x)$, then $$\frac{dy}{dx} \,=\, -\frac{\sin(x)}{1+\cos(y)},$$ so $\frac{dy}{dx}$ is not defined when $\cos(y)=-1.$

But $\cos(x)$ is differentiable at every real number, so ${y}+\sin(y)$ should also be differentiable, and so $y$ should be differentiable at every point. But it is not differentiable when $\cos(y)=-1\,$?

Answer 1

As it comes, your implicit function is differentiable at all its points.

Namely, $|\cos x|\le 1$ for all $x$. On the other hand, if you assume $\cos y=-1$ then $y$ is an odd multiple of $\pi$, so $\sin y=0$ and so $|y+\sin y|=|y|\ge\pi$. Thus, the case $\cos y=-1$ never happens on any of the points of the curve $y+\sin y=\cos x$.

The moral here is: the derivative is decided on each point on the curve, with the stress on the words on the curve.

Answer 2

Why shoud $y$ be differentiable everywhere? If $y^3=x$, then $y=\sqrt[3]x$, which is not differentiable at $0$, in spite of the fact that $y^3$ and $x$ are differentiable everywhere.

Answer 3

In the real domain the condition $\cos y=-1$ actually does not exist. It would imply that $\sin y=0$ and $y$ itself is an odd integer times $\pi$, so $|y+\sin y|=|\cos x|>1$ which cannot be true.

When $y+\sin y$ hits a bounding value of $\pm1$, the implicit derivative which has $\sin x=0$ in the numerator goes to zero, so you generate a critical point that automatically limits the range of the function to where a smooth curve is possible.

In the complex domain $|\cos x|>1$ does become possible and therefore we do have nondifferentiable points where $y=\cos x=(2n+1)\pi$. In this case what actually happens is $y+\sin y$ is no longer a bijective function of $y$, therefore the function of $y$ versus $x$ will have branch points where the derivative is undefined. Local analysis around these branch points then reveals that the branching is threefold at every such point.